Animated Solution for Mathematics - Straight Lines: The vertices of a triangle are $A(-1,3)$, $B(-2,2)$ and $C(3,-1)$. A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are embarking on a journey through the coordinate plane. We have a triangle, a simple geometric shape, but it hides a secret.

We are going to shift its boundaries, and in doing so, we will uncover the elegance of parallel lines and the power of the perpendicular distance formula. Let us begin.

Imagine you are standing on a vast, empty grid. You have three points: $A(-1,3)$, $B(-2,2)$, and $C(3,-1)$. These are the vertices of our triangle.

Before we touch a single equation, let us visualize them. $A$ is in the second quadrant, $B$ is also in the second quadrant, and $C$ is in the fourth.

When you connect these points, you see a triangle that stretches across the origin. This visualization is your most powerful tool. It tells you that the side $AC$ is the one that passes closest to the origin $(0,0)$. This is our target.

To work with side $AC$, we need its equation. We know two points: $A(-1,3)$ and $C(3,-1)$.

The slope $m$ is the change in $y$ over the change in $x$. Using the formula $m=\frac{y_2-y_1}{x_2-x_1}$, we calculate:

With a slope of $-1$ and a point $A(-1,3)$, we use the point-slope form: $y-3=-1(x-(-1))$. Expanding this, we get $y-3=-x-1$, which simplifies beautifully to:

This is the soul of our line $AC$. It is the foundation upon which we will build our solution.

Now, the problem asks us to shift the sides of the triangle one unit inwards. What does this mean? It means we are creating a new line, parallel to $x+y-2=0$, but shifted by a distance of $1$ unit towards the origin.

First, let us find the distance of the original line $AC$ from the origin. The perpendicular distance formula is $p=\frac{|c|}{\sqrt{a^2+b^2}}$. Plugging in our values, we get:

This is the distance from the origin to the original side $AC$. Since we are shifting the line inwards, towards the origin, our new distance $p^{\prime }$ will be $p-1$, which is $\sqrt{2}-1$.

We know the new line is parallel to $x+y-2=0$. Therefore, it must have the same slope. We can write its equation as $x+y-d=0$, where $d$ is the unknown constant we need to find.

We apply the perpendicular distance formula again, this time for our new line:

This simplifies to $\frac{|d|}{\sqrt{2}}=\sqrt{2}-1$. Multiplying both sides by $\sqrt{2}$, we get:

Since the line lies between the origin and the original line, $d$ must be positive. Thus, $d=2-\sqrt{2}$.

We have arrived at our destination. Substituting $d$ back into our equation, we get:

This is the equation of the side of the new triangle nearest to the origin. It matches option (3).

Look at the beauty of this result. We started with a simple triangle, used the slope to define its boundary, used the distance formula to measure its position, and used the concept of parallel lines to shift it. This is the essence of coordinate geometry: transforming a visual problem into an algebraic one, and then solving it with precision and grace.